Maths

Let $\theta$ be the angle between the vectors $\left|\overrightarrow{a}\right|$ and $\left|\overrightarrow{b}\right|$ .

It is given that $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|,\overrightarrow{a}.\overrightarrow{b}=\frac{1}{2}and\theta =60^{?}----(1)$

We know, $\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$

$\begin{array}{l}\therefore \frac{1}{2}=\left|\overrightarrow{a}\right|\left|\overrightarrow{a}\right|cos60^{?}(using(1))\\ \frac{1}{2}={\left|\overrightarrow{a}\right|}^2*\frac{1}{2}\\ {\left|\overrightarrow{a}\right|}^2=1\\ \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=1\end{array}$

$\therefore$ Magnitude of two vector=1

1. Given, f (x) = 5x 3

At x = 0,  $\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}$ 5x 3 = 5 0 3 = 3.

So f is continuous at x = 1.

At x = 3,  $\pi +h$ 5x 3 = 5 ( 3) 3 = 15 3

= 18.

So f is continuous at x = 3.

At x = 5,  $\forall x\in ?$ .5x 3 = 5.5 3 = 25 3 = 22.

So, f is continuous at x = 5.

$\left(3\overrightarrow{a}-5\overrightarrow{b}\right).\left(2\overrightarrow{a}+7\overrightarrow{b}\right).$

$\begin{array}{l}=3\overrightarrow{a}.2\overrightarrow{a}+3\overrightarrow{a}.7\overrightarrow{b}-5\overrightarrow{b}.2\overrightarrow{a}-5\overrightarrow{b}.7\overrightarrow{b}\\ =6\overrightarrow{a}.\overrightarrow{a}+21\overrightarrow{a}.\overrightarrow{b}-10\overrightarrow{a}.\overrightarrow{b}-35\overrightarrow{b}.\overrightarrow{b}\\ =6{\left|\overrightarrow{a}\right|}^2+21\overrightarrow{a}.\overrightarrow{b}-10\overrightarrow{a}.\overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2\\ =6{\left|\overrightarrow{a}\right|}^2+11\overrightarrow{a}.\overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2\end{array}$

$\left|\overrightarrow{a}\right|$ and $\left|\overrightarrow{b}\right|$ ,if $\left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)=8$ and $\left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|$

$\begin{array}{l}\left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)=8\\ and\left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|\\ \left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)=8\\ \overrightarrow{a}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}-\overrightarrow{b}.\overrightarrow{b}=8\\ {\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2=8\\ {\left(8\left|\overrightarrow{b}\right|\right)}^2-{\left|\overrightarrow{b}\right|}^2=8\\ 64{\left|\overrightarrow{b}\right|}^2-{\left|\overrightarrow{b}\right|}^2=8\\ 63{\left|\overrightarrow{b}\right|}^2=8\\ \left|\overrightarrow{b}\right|=\surd 8/\surd 63\left(\therefore magnitudeofavectorisnon-negative\right)\\ \left|\overrightarrow{b}\right|=\frac{\mathrm{2\surd 2}}{\mathrm{3\surd 7}}\\ And\\ \left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|=\frac{2*\mathrm{2\surd 2}}{\mathrm{3\surd 7}}=\frac{1\mathrm{6\surd 2}}{\mathrm{3\surd 7}}\end{array}$

47. The given system of inequality is

2x + y ≥ 4- (1)

x + y ≤ 3- (2)

2x – 3y ≤ 6- (3)

The corresponding equation are

2x + y = 4

and x + y = 3

and 2x + 3y = 6

Putting (x, y)= (0,0) in (1), (2) and (3),

2 * 0+0 ≥ 4

0 ≥ 4 which is false.

and 0+0 ≤ 3   => 0 ≤ 3 which is true.

and 2 * 0 – 3 * 0 ≤ 6  => 0 ≤ 6which is also true.

So, solution of inequality (1) excludes plane with origin while solution of inequality (2) and (3) includes the plane with origin.

Here, each of the given three vector is a unit vector.

$\begin{array}{l}\overrightarrow{a}.\overrightarrow{b}=\frac{2}{7}*\frac{3}{7}+\frac{3}{7}*\left(\frac{-6}{7}\right)+\frac{6}{7}*\frac{2}{7}\\ =\frac{6}{49}+\left(\frac{-18}{49}\right)+\frac{12}{49}=\frac{6-18+12}{49}=0\\ \overrightarrow{b}.\overrightarrow{c}=\frac{3}{7}*\frac{6}{7}+\left(\frac{-6}{7}\right)*\frac{2}{7}+\frac{2}{7}*\left(-\frac{3}{7}\right)\\ =\frac{18}{49}-\frac{12}{49}+\left(\frac{-6}{49}\right)=\frac{18-12-6}{49}=0\\ \overrightarrow{c}.\overrightarrow{a}=\frac{6}{7}*\frac{2}{7}+\frac{2}{7}*\frac{3}{7}+\left(-\frac{3}{7}\right)*\frac{6}{7}\\ =\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=0\end{array}$

Therefore, the given three vectors are mutually perpendicular to each other.

Let,

$\begin{array}{l}\overrightarrow{a}=\widehat{i}+3\widehat{j}+7\widehat{k}\\ \overrightarrow{b}=7\widehat{i}-\widehat{j}+8\widehat{k}\end{array}$

The project of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is.

46. The given system of inequality is

3x+4y ≤ 60 - (1)

x+3y ≤ 30- (2)

x≥ 0 - (3)

xy≥ 0 - (4)

The corresponding equation of (1) and (2) are

3x + 4y = 60

and x + 3y = 30

Putting (x, y)= (0,0) in equality (1) and (2),

3 * 0+4 * 0 ≤ 60 and 0+3 * 0 ≤ 30

0 ≤ 60 which is true and 0 ≤ 30 which is true

So, the solution plane of inequality (1) and (2) is the plane including origin (0,0)

∴ The shaded portion is the solution of the given system of inequality.

Let,

$\begin{array}{l}\overrightarrow{a}=\widehat{i}-\widehat{j}\\ \overrightarrow{b}=\widehat{i}+\widehat{j}\end{array}$

The projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is given by,

$\therefore$ The projection of vector $\overrightarrow{a}$ on $\overrightarrow{b}$ is 0.